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Proving the limit of a sequence using binomial theorem

Here's the question:
Capture.PNG

I worked out the limit to be 3/2. I confirmed this using wolphram alpha.
However when I try to use binomial theorem to prove this, I can't seem to get the same answer. My workings so far are:
Attachment not found


where am I going wrong?

Thanks, Charlie.
13589213_1388154607862208_1465240044_o.jpg95.2KB
Reply 1
Avatar for Zacken
Zacken
22
Original post by TurboCarlos1


where am I going wrong?

Thanks, Charlie.


Nope, nope, nope. You can't apply limits randomly like that. Not if they don't exist.

Unparseable latex formula:

\displaystyle [br]\begin{equation*} \lim_{n \to \infty} (nf(n)) \neq \bigg( \lim_{n\to \infty} n\bigg)\bigg(\lim_{n \to \infty} f(n)\bigg) \end{equation*}



since lim⁡n→∞n\lim_{n \to \infty} n doesn't exist.

Instead, observe that:

n(1+32n+O(1/n2))−n=n+32+O(1/n)−n=32+O(1/n)n\left(1 + \frac{3}{2n} + O(1/n^2)\right) - n = n + \frac{3}{2} + O(1/n) - n = \frac{3}{2} + O(1/n)

so the limiting value is 32\frac{3}{2} since all other terms are of the order n−1n^{-1} at most and go to 0 in the limit.
(edited 7 years ago)
Reply 2
Avatar for math42
math42
20
Original post by TurboCarlos1
Here's the question:
Capture.PNG

I worked out the limit to be 3/2. I confirmed this using wolphram alpha.
However when I try to use binomial theorem to prove this, I can't seem to get the same answer. My workings so far are:
Attachment not found


where am I going wrong?

Thanks, Charlie.


You're subbing in limits too hastily. Always be careful with stuff like this. You note that when the 1 + 3/2n + blah is multiplied by the n, you get n + 3/2 + (a load of stuff that tends to 0) giving you overall n + 3/2 + (a load of stuff that tends to 0) - n = 3/2 + (a load of stuff that tends to 0), which clearly tends to 3/2
thanks to you both, got it.
Reply 4
Avatar for Zacken
Zacken
22
Original post by TurboCarlos1
thanks to you both, got it.


No worries.

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